An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.70 mm.
(a) If a 16.0 V potential difference is applied to these plates, calculate the electric field between the plates.
kV/m
(b) What is the surface charge density?
nC/m2
(c) What is the capacitance?
pF
(d) Find the charge on each plate.
pC
To find the answers to the given questions, we can use the formulas and principles of capacitance and electric field.
(a) To calculate the electric field between the plates, we can use the formula:
Electric field (E) = Voltage (V) / plate separation distance (d)
Given:
Voltage (V) = 16.0 V
Plate separation distance (d) = 1.70 mm = 0.0017 m
Substituting the given values into the formula, we can calculate the electric field:
E = 16.0 V / 0.0017 m = 9411.76 V/m ≈ 9.41 kV/m
Therefore, the electric field between the plates is approximately 9.41 kV/m.
(b) To calculate the surface charge density, we can use the formula:
Surface charge density (σ) = Electric field (E) * Permittivity of free space (ε0)
The permittivity of free space is approximately ε0 = 8.85 × 10^(-12) F/m.
Substituting the given values into the formula, we can calculate the surface charge density:
σ = (9.41 kV/m) * (8.85 × 10^(-12) F/m) = 8.32 × 10^(-8) C/m² ≈ 83.2 nC/m²
Therefore, the surface charge density is approximately 83.2 nC/m².
(c) To calculate the capacitance, we can use the formula:
Capacitance (C) = (Permittivity of free space (ε0) * plate area (A)) / plate separation distance (d)
Given:
Plate area (A) = 7.60 cm² = 7.60 × 10^(-4) m²
Plate separation distance (d) = 1.70 mm = 0.0017 m
Substituting the given values into the formula, we can calculate the capacitance:
C = (8.85 × 10^(-12) F/m) * (7.60 × 10^(-4) m²) / (0.0017 m) = 3.94 × 10^(-11) F ≈ 39.4 pF
Therefore, the capacitance is approximately 39.4 pF.
(d) To find the charge on each plate, we can use the formula:
Charge (Q) = Capacitance (C) * Voltage (V)
Given:
Capacitance (C) = 39.4 pF = 39.4 × 10^(-12) F
Voltage (V) = 16.0 V
Substituting the given values into the formula, we can calculate the charge on each plate:
Q = (39.4 × 10^(-12) F) * (16.0 V) = 6.30 × 10^(-10) C ≈ 630 pC
Therefore, the charge on each plate is approximately 630 pC.